We're given 2 constants $a,b\geq 1$ and we need to calculate x such that: $x=a\log_2(bx)$.
I thought of doing the following:
$$x=\frac{ba\log_2(\frac{ba}{a}x)}{b}=\frac{ba\log_2(ba\log_2(\frac{ba}{a}x))}{b}=\frac{ba\log_2(ba\log_2(ba\log_2(...)))}{b}$$
From the second transition each time I replace $x$ with $a\log_2(bx)$ and I do it infinite times.
And then we need to find $t$ such that:
$$t=ba\log_2(t)$$ $$\frac{t}{\log_2(t)}=ba$$
And we would get that: $$x=\frac{t}{b}$$
So I have 3 question:
1) Was the first infinite log transition even "legal"?
2) If it is then how can I solve the equation for $t$? Is there any inverse function for $\frac{t}{\log_2(t)}$?
3) If the answer to 1 or 2 is no, then is there any other way to solve this?
It probably can be made legal by saying something about fixed points of function $t \to ba\log_2 t$, but we don't need it. After you got idea about $x = \frac{t}{b}$, you can substitute it to the original equation to get $t = ab\log_2 t$.
Value of $t$ can't be expressed in elementary functions, but can be in Lambert W function. Just transform your equation on $t$ to form $t = e^{ct}$ and substitute $c$ to formula there.